Inorganic notes 1st year engineering college

7/24/2019 Inorganic notes 1st year engineering college

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Kinetic stability

Inert and labile complexes


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Inert and labile complexes

Thermodynamically stable complexes can be labile or inert

The term inert and labile are relative

A good rule of thumb is that those complexes that react completely within 1 min at 25 o

should beconsidered labile and those that takelonger should beconsidered inert.

[Hg(CN)4]2-

Kf= 1042

thermodynamically stable

[Hg(CN)4]2-

+ 414

CN-

= [Hg(14

CN)4]2-

+ CN-

Very fast reaction LabileChelating agents:

(1) Used to remove unwanted metal ions in water.(2) Selective removal of Hg

2+and Pb

2+from body when poisoned.

(3) Prevent blood clots.

(4) Solubilize iron in plant fertilizer.

2,3-dimercapto-1-propanesulfonic acid sodium (DMPS) DMPS is a effective chelator with two groups thiols - for mercury, lead,tin, arsenic, silver and cadmium.

Important Chelating Ligands

HO

OSH Zn AsH


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HOOH

O SH

(R,S )-2,3-dimercaptosuccinic acid

As, Cu, Pb, Hg

HS OH

SH

M+

S

SOH

M

Dimercaprol

AsHg

AuPb

D-Penicillamine

Hg AuPb

Ca 2+EDTA: another view

Important Chelating Lig


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Anticoagulant Macrocylic Ligands


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Homogeneous catalysis

Cat al yzedrxnCatalyst


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Important catalyst properties

* Activity: A reasonable rate of reaction is needed

Turnover frequency (N)

N = /[Q]

Large turnover frequency efficient catalyst

* Selectivity: Byproducts should be minimized

* Lifetime: It is costly to replace the catalyst frequently

* Cost: The acceptable cost depends upon the catalyst lifetimand product value

G

GRea ctan ts

Product

E a

E acata ly zed

Cat al yzed rxnproceeding through

an i ntermedi at e A + B C

Heterogeneous

Homogeneous

A catalyst lower the activation barrier for a transformation, byintroducing a new reaction pathway.

It does not change the thermodynamics!!

Transition metal organometallic compounds

Metal-carbon bond

Organic compounds Octet rule

Organometallic 18 electron rule


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18 en rule

* 18 valance electron inert gas configuration

Oxidation state method Neutral atom method

Ligand Name Bonding Type FormalCharge

Electronsdonated

Molecular Hydrogen:H2

0 2

Hydride H - M -H -1 2

Halide X - M-X -1 2

Amine, phosphine,arsine: NR3, PR3, AsR3

M -NR3 M -PR3 0 2

Carbonyl: C O 0 2

Alkyl , Aryl M-CR M-Ph -1 2

Alkene -1 2

MH

H

CM O

CHH C

M

FeOC CO

COFe is 4s23d6 = 8e

0

RhPh3P ClPh3P PPh3 Rh is s1d8 = 9e

i Cl i 1 Rh i


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OCCO

each Co is neutral so Fe 0

each CO donates 2 e = 10e

8e + 10e = 18ecoordinately saturated

since Cl is -1, Rh is

4 ligands x 2e each =9e - 1e + 8e = 16e

therefore coordinately unsaturated

16 electron complexes

Group 9 and 10

IrCl

PPh 3

Ph 3P

OC

9+1+2+2+2 = 16

Exception to 16/18 electron rule

V(CO)6 17 electrons

W(CH3) 12 electrons

Catalytic steps

(a) Ligand coordination and dissociation

Facile coordination of the reactant and facile loss of products.

Coordinatively unsaturated - 16-electron complexes

(b) Oxidative addition

Non-bonding electron pair in the metal

Coordinatively unsaturated

Oxidation of metal by two units Mn

to Mn+2

Ph 3P Cl RI Ph 3P ClR Oxidative addition (c) Insertion or migration

Migration of alkyl and hydride ligands


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IrCOPh 3P

+ RI IrCO

3

Ph 3P I

Migration of alkyl and hydride ligands

L + M CO

R

MH CH 2

CH 2

(d) Nucleophilic attack

C

C

R R

H R

L 3Pd OH 2 CR

H

C

R

R

2+

L3Pd

L 5M CO + OH - L 5M C

O

O

(e) Reductive elimination

Involves decrease in the oxidation and coordinationnumber

Rh

COPh 3P

Ph 3P Me

COR

Cl

RhCOPh 3P

Ph 3P ClRCOMe+

Wilkinsons CatalystTris(triphenylphosphine)rhodium(I) chloride

Synthesis


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Rh

Ph 3 P PPh 3

PPh 3Cl

Ph 3P

Cl

EtOH

78 oC

+RhCl3

3 H2O + >4 PPh

3

commercially available Sigma Aldrich, Acro

RhPh3P Cl

Ph3P PPh 3+ H2 Rh

Ph3P

Cl

Ph3PH

RhPh3P

Cl

Ph3P PPh 3H

HRh

Ph3PCl

Ph3PH

H + PP

(1) Oxidative addition

(2) Ligand Dissociation

Ph3PH

CH 2 Ph3PH(3) Ligand Association


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RhPh3P

Cl

3H

2

CH 2Rh

Ph3PCl

Ph3P+

RhPh3P

Cl

Ph3P H

H

CH2

CH 2Rh

Ph3PCl

Ph3PCH

CHH(4) Insertion

+ PPh 3RhPh3P

Cl

Ph3PCH 2

H

CH 2H

RhPh3P

Cl

Ph3PCH

CHH(5) Ligand association

Ph PCH 2CH 2H

PPh

(6) Reductive elimination


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RhPh 3 P Cl

Ph 3 P PPh 3RhPh 3 P

Cl

Ph 3 P

H

PPh 3

(note: regeneration of the catalyst)

WCINATION


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Highly sensitive to the nature of the phosphine

Analogous complexes with alkylphosphine

Chiral phosphine ligands have been developed optically active products.

Analogous complexes with alkylphosphin

Applications * Laboratory scale organic synthesis* Production of fine chemicals* Synthesis of L-DOPA

Used for the treatment of Parkinsons diseases

Synthetic route was developed by Knowles and co-workers at Monsanto

This reaction, developed by Knand Sabacky, was used at M


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com mer ci al r ou te to th e Par kDOPA.

M etal Carbonyl Compl exes

M-CO

CO is an i nert m olecule that becomes activated bycomplexation to metals

CO as a ligandstrong donor, strong -acceptor

strong trans effectsmall steric effect

Homolepti

Group Formula Valence en Structure

6 Cr(CO)6 6 +12 = 18 COCO


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7 Mn2(CO)10 7+10+1 = 18

8 Fe(CO)5 8+10 = 18

9 Co2(CO) 9+8+1= 18

10 Ni(CO)4 10+8 = 18

Cr

COCO

OC

OC

Mn Mn

OC

CO

OC CO

CO

CO

CO

OC

CO

OC

FeCO

CO

COOC

OC

Co Co

CO

CO

OCCO

OC

CO

CO

OC

Ni

CO

COCOOC


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O C M

orbital serves as a very weak donor to a metal atom

O C M O C M O C M

CO-M sigma bond M to CO pi backbonding CO to M pi bonding (rare)

Metal should be in low oxidation state ie en rich

SynthesisDirect combination

Ni( ) 4CO( ) Ni(CO) (l)

Substituted carbon

Oxidative Addition


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Ni(s) + 4CO(g) Ni(CO) 4(l)Reductive carbonylation 30

oC and 1 atm CO

CrCl 3(s) + Al (s) + 6CO (g) AlCl 3(soln) + Cr(CO) 6 (soln)

AlCl3 and Benzene

Re2O7(s) + 17 CO (g) Re 2(CO) 10(s) + 7CO 2(g)250

oC, 350 atm CO

Concomitant oxidation of the metal and addition of

Fe(CO)5 + Br2 Fe(CO)4Br2 + CO

Fe0

Fe2+

Reductive Carbonylation

RhCl 3.3H 2O RCO/EtOH

100 o

Metal halide and CO

Properties

Reduction to form metal carbonylates

Fe(CO) 5 [Fe(CO) 4]2- + CO Na, THFSubstitution

Cr(CO) 5(solv) +PR 3 Cr(CO) 5PR 3 + solvCr(CO) 6 + Solv Cr(CO) 5(solv) + CO

[Cr(CO) 5Solv] Cr(CO) 5(PR 3) - C

OSo l v


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Cr(CO) 6

Cr(CO) 6(PR 3)]# [Cr(CO) 5(PR 3)]

+ S o

+ P R 3 -CO

Oxidation

(CO) 5Mn-Mn(CO) 5 + Br 2 2Mn(CO) 5BrOxidative cleavage of M-M bond

Protonation

Mn(CO) 5-(aq) + H +(aq) HMn(CO) 5(s)

more negative the anion, higher its Bronsted basicity

use in the synthesis of different organometallic

[Mn(CO) 5]- + CH 3I [Mn(CH 3)(CO) 5] + I

-

ApplicationHydroformylation


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Co 2(CO) 8

HCo(CO) 4

CO

HCo(CO) 3

CH 2=CHR

H(CO) 3Co

(CO) 3CoCH 2CH(CO) 4CoCH 2CH 2R

CO

(CO) 3Co C-CH 2CH 2R

RCH 2CH 2CHO

H 22HCo(CO) 4

Biology


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Biology


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0

Industry Synthesis of different compoundsExtraction of elements


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Environment


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Balancing Redox Equations

1. Assign oxidation numbers to each atom. Determine the

Fe 2+ + MnO 4- + H + Mn 2+ + Fe 3+ + H 2OExample

Fe 2+ + MnO 4- + H +

2


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elements that get oxidized and reduced. Split the equationinto half-reactions.

2. Balance all atoms in each half-reaction, except H and O.3. Balance O atoms using H 2O. Balance H atoms using H +.

7. Balance charge using electrons.8. Sum together the two half-reactions, so that: e - lost = e -

gained9. If the solution is basic, add a number of OH - ions to each

side of the equation equal to the number of H + ions shownin the overall equation. Note that H + + OH - H 2O

MnO 4- Mn 2+ Red(+7) (+2)

Fe 2+ Fe 3+ Oxid

MnO 4- + 8H + + 5e Mn

5Fe 2+ 5Fe 3+ +5e5Fe 2+ + MnO 4- + 8H +

Nernst EquationaOx 1 +bRed 2 a

Q =[Red 1]a [Ox 2]b

[Ox 1]a [Red 2]bE =

E 0 = Standard PotentialR = Gas constant 8.314 J/K.mF- Faraday constant = 94485 Jn- number of electrons

G 0 = - n F E 0

Note: if G 0 < 0, then E 0 must be >0A reaction is favorable if E 0 > 0

2H + (aq) + 2e H 2(g)E 0 (H +, H 2) = 0

Zn 2+ (aq) + 2e Zn(s)E 0 (Zn 2+ , Zn) = -0.76 V

2H + (aq) + Zn(s) Zn 2+(aq) + H 2(g)E 0 = +0.76 V

Hydrogen Electrode consists of a platinum electrode covered

with a fine powder of platinum around


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with a fine powder of platinum aroundwhich H 2(g) is bubbled. Its potential isdefined as zero volts.

Hydrogen Half-Cell

H2(g) = 2 H+

(aq) + 2 e-

reversible reaction

Galvanic

Cell


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-

Latimer Diagram* Written with the most oxidized species on the left, and the most

reduced species on the right


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* Oxidation number decrease from left to right and the E 0 valuesare written above the line joining the species involved in thecouple.

reduced species on the right.

A+5 B+3 C +1 D0 E -2x y zw

Iron +2 and +3What happens when Fe(s) react with H +?

G = -nF


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Fe 2+ + 2e Fe

Fe3+

+ e Fe2+

Fe 3+ + 3e Fe

-2 x F x -0.44 = 0.8

-1 x F x +0.771 = -0.7

+ 0.109 F= -3 x F x 0.036

Fe

Fe3++0.036

Fe2++0.44

Fe 3+ Fe 2+ Fe

-0.036

+0.77 -0.44

-0.440

-0.771

-0.036


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Fe 3+ Fe 2++0.77 -0.44 Fe

[Fe(CN) 6]3- [Fe(CN) 6]4--1.160.36

Oxidation of Fe(0) to Fe(II) is considerably more favorable in thecyanide/acid mixture than in aqueous acid.

(1) Concentration(2) Temperature(3) Other reagents which are not inert


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Oxidation of elemental copper

Latimer diagram for chlorine in acidic solution

+1 2 +1.18 +1 65 +1.63 +1 3


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ClO 4- ClO 3

- HClO 2 HClO Cl 2 +1.2 1.18 +1.65 1.63 +1.3

+7 +5 +3 +1 0

ClO4

- ClO3

-+1.2

HClO Cl 2+1.63

2 HClO(aq) + 2 H +(aq) + 2 e - Cl 2(g) + 2 H 2O(l) E 0 = +1.6

Can you balance the equation?

balance the equation

How to extract E 0 for nonadjacent oxidation state?E

0 =?


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ClO 4- ClO 3

- HClO 2 HClO Cl 2 +1.2 +1.18 +1.65 +1.63 +1.3

+7 +5 +3 +1 0

Write the balanced equation for the first couple

Write the balanced equation for the second couple

1

HClO(aq) + H +(aq) + e Cl 2(g) + H 2O(l) +1.63 V

Cl 2(g) + e Cl - (l) +1.36 V

G = G + G

- FE = - FE - E = E+ E

+

Find out the oxidation state of chlorine

E 1 5 V

Identify the two reodx couples

0+1

Latimer diagram for chlorine in basic solution

+1 2 +1.18 +1.65 +1.63 +1.3


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ClO 4-

ClO 3-

ClO 2-

ClO-

Cl 2+0.37 +0.3 +0.68 +0.42 +1.36

+0.89

ClO - Cl2+0.42

2ClO - (aq) + 2H 2O(l) + 2e - Cl2(g) + 4OH -(aq) E 0 = 0.42

ClO 4- ClO 3

- HClO 2 HClO Cl 2 +1.2

+7 +5 +3 +1 0

+0.89

Balance the equation

Find out the E 0

+0.89

DisproportionationElement is simultaneously oxidized and reduced.


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ClO 4- ClO 3- ClO 2- ClO - Cl 2+0.37 +0.3 +0.68 +0.42 +1.36

2M +(aq) M2+ (aq)M (s)E 0 E 0

2 M +(aq) M(s) + M 2+(aq)

the potential on the left of a species is less positive than that on

the right- the species can oxidize and reduce itself, a process kas disproportionation .

ClO - Cl 2 Cl -+0.42 +1.36

ClO 4- ClO 3- ClO 2- ClO - Cl 2+0.37 +0.3 +0.68 +0.42 +1.36


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Cl2 + 2OH - ClO - + Cl - + H 2O

ClO - Cl 2 Cl -+0.42 +1.36

E = E 0 (Cl 2/Cl -) E0 (ClO -/Cl 2) = 1.36 - +0.42 = 0.94

Reaction is spontaneous

Cl2(g) + 2 e - 2Cl -(aq) +1.362ClO - (aq) 2H 2O(l) +2e - Cl2(g) + 4OH -(aq) +0

Latimer diagram for Oxygen


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1.23 V

the potential on the left of a species is less positive than that on theright-

the

species

can

oxidize

and

reduce

itself

a

process

know

Disproportionation


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right- the species can oxidize and reduce itself, a process knowas disproportionation .

Is it spontaneous


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H2O2(aq) + 2H + (aq) +2e - 2H 2O(aq) +1.76 V

O2(g) + 2H +(aq) +2 e - H

2O

2(aq) +0.7 V

H2O2(aq) O 2 (g) + H 2O(l) +0.7 V

Yes the reaction is spontaneous


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Another example


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2 Cu +(aq) Cu 2+(aq) + Cu(s)

Cu +(aq) + e - Cu(s) E 0 = + 0.52 V

Cu 2+(aq) + e - Cu +(aq) E 0 = =0.16 V

Cu(I ) undergo dispropor tionation in aqueous solu tion

Comproportionation reaction


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Ag 2+(aq) + Ag(s) 2Ag +(aq) E 0 = + 1.18 V

Reverse of disproportionation

we will study this in detail under Frost diagram

Frost DiagramGraphically illustration of the stability of different oxidationstates relative to its elemental form (ie, relative to oxidat

Arthur A. Frost


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(state= 0 )

X N + Ne -

NE 0 = -G 0

Look at the Latimer diagram of nitrogen in acidic solution


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a b c d e

f g h

ac


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N2

b

d

e

f

g

h

G = G + G-nFE = -n FE -E = nE+ nE

n+n

NO3

- + 6H + + 5e - N2

+ 3H2O E 0 = 1.25V

N 2O4 + 4H + + 4e - N 2 + 2H 2O E 0 = 1.36V

a

b


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HNO 2 + 3H + + 3e - N 2 + 2H 2O E 0 = 1.45V

NO + 2H+

+ 2e-

N 2 + H 2O E0

= 1.68V

N 2O + H + + e - N 2 + H 2O E 0 = 1.77V

N 2 + 2H + + H 2O + e - NH 3OH + E0 = -1.87V

N 2 + 5/2 H+

+ 2e-

N 2H5+

E0

= -0.23V

N 2 + 4H + + 3e - NH 4+ E0 = 0.27V

c

d

e

f

g

h

N(V): NO 3- (5 x 1.25, 5)

Oxidation state: species NE 0, N


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N(IV): N 2O4 (4 x 1.36, 4)

N(III): HNO 2 (3 x 1.35, 3)

N(II): NO (2 x 1.68, 2)

N(I): N 2O (1 x 1.77, 1)

N(-I): NH 3OH + [-1 x (-1.87), -1]

N(-II): N 2H5+ [-2 x (-0.23), -2]

N(-III): NH 4+ (-3 x 0.27, -3)

Frost Diagram

N

2


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What do we really get from the Frost diagram?


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the lowest lying species corresponthe most stable oxidation state of element

Slope

of

the

line

joining

any

two

points

is

equal

to

the

spotential

of

the

couple

.


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N

NE 0

NE 0

N Slope = E 0= NE 0

N -

E

0

of a redox coupleHNO

2

/NO


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3, 4.4

2, 3.4

Slope = E 0= NE0

N -

1 V

Th idi i l i h i i l

Oxidizing agent? Reducing agent?


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The oxidizing agent - couple with more positive slope - morepositive E

The reducing agent - couple with less positive slope

If the line has ive slope- higher lying species reducing agent

If the line has +ive slope higher lying species oxidizing agen

Identifying strong or weak agent?


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NO Strong oxidant than HNO 3


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DisproportionationElement is simultaneously oxidized and reduced.

2 M +(aq) M(s) + M 2+(aq)


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2M +(aq) M2+ (aq)M (s)E 0 E 0

the potential on the left of a species is less positive than that onthe right- the species can oxidize and reduce itself, a process knas disproportionation .

DisproportionationWhat Frost diagram tells about this reaction?


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A

species

in

a

Frost

diagram

is

unstable

with

respect

to

disproportionationif

its

point

lies

above

the

line

connecting

two

adjacent

species

.


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Disproportionation . another example


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Comproportionation is spontaneous if the intermediate specieslies below the straight line joining the two reactant species.


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Favorable ?


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NE 0

Disproportionation


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Comproportionation

In acidic solution


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Mn and MnO 2

Mn 2+

Rate of the reaction hinderedinsolubility?

In basic solution

MnO 2 and Mn(OH) 2

Mn 2O 3

* Thermodynamic stability is found at the bottom of the diagram.Mn (II) is the most stable species.

From the Frost diagram for Mn


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* A species located on a convex curve can undergo disproportionation

example: M nO 4 3-

M nO 2 and M nO 4 2-

(in basic solu tion)

Any species located on the upper right side of the diagram will be

a strong oxidizing agent. MnO 4-

- strong oxidizing agent.

Any species located on the upper left side of the diagram will bea reducing agent. Mn - moderate reducing agent .

* Although it is thermodynamically favorable for permanganateion to be reduced to Mn(II) ion, the reaction is slow except in thepresence of a catalyst. Thus, solutions of permanganate can bestored and used in the laboratory


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* Changes in pH may change the relative stabilities of the species.The potential of any process involving the hydrogen ion willchange with pH because the concentration of this species ischanging.

* Under basic conditions aqueous Mn 2+ does not exist. InsteadInsoluble Mn(OH) 2 forms.

stored and used in the laboratory.

*All metals are good reducing agents

*Exception: Cu*Reducing strength: goes downsmoothly from Ca to Ni


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smoothly from Ca to Ni*Ni- mild reducing agent

*Early transition elements: +3 stateLatter +2 state

*Fe and Mn many oxidation stat*High oxidation state:

Strong oxidizing agents

Inorganic notes 1st year engineering college

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